Question

In Dumas' method for estimation of nitrogen 0.4 g of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound is (Given : Aqueous tension at 300 K = 15 mm Hg)

• A. 15.71%
• B. 20.95%
• C. 17.46%
• D. 7.85%

Hint:

In Dumas' method, remember to subtract the aqueous tension from the total pressure to get the pressure of dry nitrogen gas. Then use the ideal gas law to find the moles of nitrogen evolved. Finally, calculate the mass of nitrogen and its percentage in the organic compound.

Answer 1

The Correct Option is A

Given: Mass of organic compound = 0.4 g 
Volume of nitrogen collected = 60 mL = \( 60 \times 10^{-3} \) L 
Temperature (T) = 300 K Pressure of wet nitrogen = 715 mm Hg 
Aqueous tension (partial pressure of water vapor) = 15 mm Hg 
The pressure of dry nitrogen (P) is: \[ P = \text{Pressure of wet nitrogen} - \text{Aqueous tension} \] \[ P = 715 \, \text{mm Hg} - 15 \, \text{mm Hg} = 700 \, \text{mm Hg} \] To use the ideal gas law (PV = nRT), we need to convert the pressure to atm: \[ P (\text{atm}) = \frac{700}{760} \, \text{atm} \] The ideal gas constant R = 0.0821 L atm mol\(^{-1}\) K\(^{-1}\). Now, we can calculate the number of moles (n) of nitrogen gas evolved using the ideal gas law: \[ n = \frac{PV}{RT} = \frac{\left( \frac{700}{760} \right) \times (60 \times 10^{-3})}{0.0821 \times 300} \] \[ n = \frac{42000 \times 10^{-3}}{760 \times 0.0821 \times 300} = \frac{42}{760 \times 24.63} = \frac{42}{18718.8} \approx 0.002244 \, \text{moles} \] The molar mass of nitrogen gas (N\( _2 \)) is 28 g/mol. The mass of nitrogen evolved is: \[ \text{Mass of N}_2 = n \times \text{Molar mass of N}_2 = 0.002244 \, \text{moles} \times 28 \, \text{g/mol} \approx 0.06283 \, \text{g} \] The percentage composition of nitrogen in the organic compound is: \[ % \text{ Nitrogen} = \frac{\text{Mass of nitrogen}}{\text{Mass of organic compound}} \times 100 \] \[ % \text{ Nitrogen} = \frac{0.06283 \, \text{g}}{0.4 \, \text{g}} \times 100 \approx 15.7075 % \] Rounding to two decimal places, the percentage composition of nitrogen is 15.71%.